volume of regular octahedron circumscribed about sphere of radius 1

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volume of regular octahedron circumscribed about sphere of radius 1

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Source: Mexican Mathematical Olympiad 1988 OMM P8

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#1 h Jul 27, 2018, 9:59 PM • 1 Y

Y by Adventure10

Compute the volume of a regular octahedron circumscribed about a sphere of radius $1$ .

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#2 h Oct 6, 2021, 7:34 PM

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It is well known that the volume of a regular octahedron with side $x$ is $\frac{\sqrt{2}}{3}x^3$ .

The octahedron will be pointing upwards, such that when we divide it in a top and bottom half, we obtain two identical pyramids with a square base. Let $A$ be the highest point in the octahedron. Let $O$ be the center of the octahedron (it is also the center of the square formed by its vertices in the middle). We will take 2 opposite sides of the octagon on its top half. Let $P$ and $Q$ be the feet of the altitude from $O$ to said sides. The sphere of radius 1 will be tangent to the sides we chose at $P$ and $Q$ , so $OP=OQ=1$ . Now, we will create a 2D diagram of the figure by slicing the octahedron with a plane passing through $O, P, Q$ . This plane will also pass through $A$ . Let $B$ and $C$ be the points at which the plane slices the sides of the middle square. Also $B$ and $P$ are on the same side wrt. line $AO$ .

We know $OP\perp AB$ , $OQ\perp AC$ , and $AB=AC$ , so $AO\perp BC$ . Then $\frac{AO}{OP}=\frac{AB}{BO}$ . When substituting these lengths for their value in terms of $x$ , we obtain $\frac{\frac{x}{\sqrt{2}}}{1}=\frac{\frac{\sqrt{3}}{2}x}{\frac x 2} \implies x = \sqrt 6$ , so the volume of the octahedron is $\frac{\sqrt{2}}{3}\sqrt{6}^3 = 4\sqrt{3}$ .

This post has been edited 2 times. Last edited by JuanDelPan, Oct 6, 2021, 7:41 PM

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